JEE Advanced 2025 Paper 2Question 1
For a non-zero complex number $z$, let $\arg(z)$ denote the principal argument of $z$, with $-\pi<\arg(z)\le\pi$. Let $\omega$ be the cube root of unity for which $0<\arg(\omega)<\pi$. Let $$\alpha=\arg\left(\sum_{n=1}^{2025}(-\omega)^n\right).$$ Then the value of $\frac{3\alpha}{\pi}$ is ______.
Complex Numbers
Answer$-2$
Solution
$-\omega=e^{-i\pi/3}$ has order $6$, and $2025=6\cdot337+3$, so the sum is $x+x^2+x^3$ with $x=e^{-i\pi/3}$: $-i\sqrt3-1$. Its argument is $\alpha=-\frac{2\pi}3$, so $\frac{3\alpha}\pi=-2$.